Mixing
Practical aspect of diffeomorphisms
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I've been trying to learn differential geometry (for awhile now), and one thing that has confused me a little is what it means from a practical perspective to have a diffeomorphism between two manifolds.
I understand the basic concept of equivalence: for instance, in algebra, when I hear: "These two groups are isomorphic," I have no problem thinking of them as exactly the same thing up to a (superficial) renaming of the elements.
And in topology, when I hear: "These two spaces are homeomorphic," I have no problem visualizing one space to be exactly the same as the other up to some (superficial) stretching or twisting.
However, in differential geometry, when I hear: "These manifolds are diffoemorphic," I'm not really sure how to think about them and the situation seems less straightforward to me. Maybe this is because I'm associating a lot of additional things to a manifold which do not come purely from it's smooth structure. At the same time, given two diffoemorphic manifolds, assuming one of them has some additional structures (e.g., tensor field, differential form, complex structure), then the other manifold also carries corresponding structures induced from the diffeomorphism.
To be more specific, let's say $M$ and $N$ are two manifolds that are diffeomorphic ($phi: M rightarrow N$) to one another. And, for example, maybe I'm given manifold $M$, but I'd prefer to think about it as $N$ because I'm more comfortable with $N$.
My impression is that I can now do anything on $N$ that depends only on the smooth structure - for example, construct differential forms, tensor fields, and vector fields - and, given my diffeomorphism, view these constructions as also existing on $M$. (this is because the tangent and cotangent bundles, and all tensor products of these are also diffeomorphic/isomorphic).
e.g.
1) If $h$ is a metric on $N$, $g = h circ dphi$ is a metric on $M$.
2) If $J$ is a complex structure on $N$, $J' = dphi^{-1} circ J circ d phi$ is a complex structure on $M$.
3) If $g$ and $h$ are as from 1), then $int_N dV_h = int_M dV_g$.
4) Any curvature at a point $p in M$ will be equal to the same curvature at the corresponding point $phi(p)in N$.
Is this correct/does this capture the essence of what it means to be diffeomorphic?
Is it even necessary to refer back to $M$? Or in practice once you have a diffeomorphism can you just work on $N$ and the fact that everything carries over to $M$ is obvious/implicit?
Are there other or better ways to think about diffeomorphism?
differential-geometry manifolds smooth-manifolds
asked Dec 22 '16 at 20:53
AshleyAshley
1,148715
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add a comment |
$begingroup$
I've been trying to learn differential geometry (for awhile now), and one thing that has confused me a little is what it means from a practical perspective to have a diffeomorphism between two manifolds.
I understand the basic concept of equivalence: for instance, in algebra, when I hear: "These two groups are isomorphic," I have no problem thinking of them as exactly the same thing up to a (superficial) renaming of the elements.
And in topology, when I hear: "These two spaces are homeomorphic," I have no problem visualizing one space to be exactly the same as the other up to some (superficial) stretching or twisting.
However, in differential geometry, when I hear: "These manifolds are diffoemorphic," I'm not really sure how to think about them and the situation seems less straightforward to me. Maybe this is because I'm associating a lot of additional things to a manifold which do not come purely from it's smooth structure. At the same time, given two diffoemorphic manifolds, assuming one of them has some additional structures (e.g., tensor field, differential form, complex structure), then the other manifold also carries corresponding structures induced from the diffeomorphism.
To be more specific, let's say $M$ and $N$ are two manifolds that are diffeomorphic ($phi: M rightarrow N$) to one another. And, for example, maybe I'm given manifold $M$, but I'd prefer to think about it as $N$ because I'm more comfortable with $N$.
My impression is that I can now do anything on $N$ that depends only on the smooth structure - for example, construct differential forms, tensor fields, and vector fields - and, given my diffeomorphism, view these constructions as also existing on $M$. (this is because the tangent and cotangent bundles, and all tensor products of these are also diffeomorphic/isomorphic).
e.g.
1) If $h$ is a metric on $N$, $g = h circ dphi$ is a metric on $M$.
2) If $J$ is a complex structure on $N$, $J' = dphi^{-1} circ J circ d phi$ is a complex structure on $M$.
3) If $g$ and $h$ are as from 1), then $int_N dV_h = int_M dV_g$.
4) Any curvature at a point $p in M$ will be equal to the same curvature at the corresponding point $phi(p)in N$.
Is this correct/does this capture the essence of what it means to be diffeomorphic?
Is it even necessary to refer back to $M$? Or in practice once you have a diffeomorphism can you just work on $N$ and the fact that everything carries over to $M$ is obvious/implicit?
Are there other or better ways to think about diffeomorphism?
differential-geometry manifolds smooth-manifolds
asked Dec 22 '16 at 20:53
AshleyAshley
1,148715
$endgroup$
1
$begingroup$
To see you write that you "have no trouble thinking" of two isomorphic groups as "exactly the same thing up to a (superficial) renaming of the elements" tells me that you misunderstand the nature of isomorphism. The renaming is far from superficial, given that it is not unique. For example, the automorphism group of $mathbb{R}^2$ under addition is uncountably infinite, and if $G$ is isomorphic to $mathbb{R}^2$ then you can precompose any isomorphism $mathbb{R}^2 mapsto G$ with any automorphism of $mathbb{R}^2$ to get a different isomorphism $mathbb{R}^2 mapsto G$.
$endgroup$
– Lee Mosher
Dec 23 '16 at 0:08
add a comment |
$begingroup$
I've been trying to learn differential geometry (for awhile now), and one thing that has confused me a little is what it means from a practical perspective to have a diffeomorphism between two manifolds.
I understand the basic concept of equivalence: for instance, in algebra, when I hear: "These two groups are isomorphic," I have no problem thinking of them as exactly the same thing up to a (superficial) renaming of the elements.
And in topology, when I hear: "These two spaces are homeomorphic," I have no problem visualizing one space to be exactly the same as the other up to some (superficial) stretching or twisting.
However, in differential geometry, when I hear: "These manifolds are diffoemorphic," I'm not really sure how to think about them and the situation seems less straightforward to me. Maybe this is because I'm associating a lot of additional things to a manifold which do not come purely from it's smooth structure. At the same time, given two diffoemorphic manifolds, assuming one of them has some additional structures (e.g., tensor field, differential form, complex structure), then the other manifold also carries corresponding structures induced from the diffeomorphism.
To be more specific, let's say $M$ and $N$ are two manifolds that are diffeomorphic ($phi: M rightarrow N$) to one another. And, for example, maybe I'm given manifold $M$, but I'd prefer to think about it as $N$ because I'm more comfortable with $N$.
My impression is that I can now do anything on $N$ that depends only on the smooth structure - for example, construct differential forms, tensor fields, and vector fields - and, given my diffeomorphism, view these constructions as also existing on $M$. (this is because the tangent and cotangent bundles, and all tensor products of these are also diffeomorphic/isomorphic).
e.g.
1) If $h$ is a metric on $N$, $g = h circ dphi$ is a metric on $M$.
2) If $J$ is a complex structure on $N$, $J' = dphi^{-1} circ J circ d phi$ is a complex structure on $M$.
3) If $g$ and $h$ are as from 1), then $int_N dV_h = int_M dV_g$.
4) Any curvature at a point $p in M$ will be equal to the same curvature at the corresponding point $phi(p)in N$.
Is this correct/does this capture the essence of what it means to be diffeomorphic?
Is it even necessary to refer back to $M$? Or in practice once you have a diffeomorphism can you just work on $N$ and the fact that everything carries over to $M$ is obvious/implicit?
Are there other or better ways to think about diffeomorphism?
differential-geometry manifolds smooth-manifolds
asked Dec 22 '16 at 20:53
AshleyAshley
1,148715
$endgroup$
I've been trying to learn differential geometry (for awhile now), and one thing that has confused me a little is what it means from a practical perspective to have a diffeomorphism between two manifolds.
I understand the basic concept of equivalence: for instance, in algebra, when I hear: "These two groups are isomorphic," I have no problem thinking of them as exactly the same thing up to a (superficial) renaming of the elements.
And in topology, when I hear: "These two spaces are homeomorphic," I have no problem visualizing one space to be exactly the same as the other up to some (superficial) stretching or twisting.
However, in differential geometry, when I hear: "These manifolds are diffoemorphic," I'm not really sure how to think about them and the situation seems less straightforward to me. Maybe this is because I'm associating a lot of additional things to a manifold which do not come purely from it's smooth structure. At the same time, given two diffoemorphic manifolds, assuming one of them has some additional structures (e.g., tensor field, differential form, complex structure), then the other manifold also carries corresponding structures induced from the diffeomorphism.
To be more specific, let's say $M$ and $N$ are two manifolds that are diffeomorphic ($phi: M rightarrow N$) to one another. And, for example, maybe I'm given manifold $M$, but I'd prefer to think about it as $N$ because I'm more comfortable with $N$.
My impression is that I can now do anything on $N$ that depends only on the smooth structure - for example, construct differential forms, tensor fields, and vector fields - and, given my diffeomorphism, view these constructions as also existing on $M$. (this is because the tangent and cotangent bundles, and all tensor products of these are also diffeomorphic/isomorphic).
e.g.
1) If $h$ is a metric on $N$, $g = h circ dphi$ is a metric on $M$.
2) If $J$ is a complex structure on $N$, $J' = dphi^{-1} circ J circ d phi$ is a complex structure on $M$.
3) If $g$ and $h$ are as from 1), then $int_N dV_h = int_M dV_g$.
4) Any curvature at a point $p in M$ will be equal to the same curvature at the corresponding point $phi(p)in N$.
Is this correct/does this capture the essence of what it means to be diffeomorphic?
Is it even necessary to refer back to $M$? Or in practice once you have a diffeomorphism can you just work on $N$ and the fact that everything carries over to $M$ is obvious/implicit?
Are there other or better ways to think about diffeomorphism?
differential-geometry manifolds smooth-manifolds
differential-geometry manifolds smooth-manifolds
asked Dec 22 '16 at 20:53
AshleyAshley
1,148715
asked Dec 22 '16 at 20:53
AshleyAshley
1,148715
asked Dec 22 '16 at 20:53
AshleyAshley
1,148715
asked Dec 22 '16 at 20:53
AshleyAshley
1,148715
asked Dec 22 '16 at 20:53
AshleyAshley
1,148715
1,148715
1
$begingroup$
To see you write that you "have no trouble thinking" of two isomorphic groups as "exactly the same thing up to a (superficial) renaming of the elements" tells me that you misunderstand the nature of isomorphism. The renaming is far from superficial, given that it is not unique. For example, the automorphism group of $mathbb{R}^2$ under addition is uncountably infinite, and if $G$ is isomorphic to $mathbb{R}^2$ then you can precompose any isomorphism $mathbb{R}^2 mapsto G$ with any automorphism of $mathbb{R}^2$ to get a different isomorphism $mathbb{R}^2 mapsto G$.
$endgroup$
– Lee Mosher
Dec 23 '16 at 0:08
add a comment |
1
$begingroup$
To see you write that you "have no trouble thinking" of two isomorphic groups as "exactly the same thing up to a (superficial) renaming of the elements" tells me that you misunderstand the nature of isomorphism. The renaming is far from superficial, given that it is not unique. For example, the automorphism group of $mathbb{R}^2$ under addition is uncountably infinite, and if $G$ is isomorphic to $mathbb{R}^2$ then you can precompose any isomorphism $mathbb{R}^2 mapsto G$ with any automorphism of $mathbb{R}^2$ to get a different isomorphism $mathbb{R}^2 mapsto G$.
$endgroup$
– Lee Mosher
Dec 23 '16 at 0:08
1
1
$begingroup$
To see you write that you "have no trouble thinking" of two isomorphic groups as "exactly the same thing up to a (superficial) renaming of the elements" tells me that you misunderstand the nature of isomorphism. The renaming is far from superficial, given that it is not unique. For example, the automorphism group of $mathbb{R}^2$ under addition is uncountably infinite, and if $G$ is isomorphic to $mathbb{R}^2$ then you can precompose any isomorphism $mathbb{R}^2 mapsto G$ with any automorphism of $mathbb{R}^2$ to get a different isomorphism $mathbb{R}^2 mapsto G$.
$endgroup$
– Lee Mosher
Dec 23 '16 at 0:08
$begingroup$
To see you write that you "have no trouble thinking" of two isomorphic groups as "exactly the same thing up to a (superficial) renaming of the elements" tells me that you misunderstand the nature of isomorphism. The renaming is far from superficial, given that it is not unique. For example, the automorphism group of $mathbb{R}^2$ under addition is uncountably infinite, and if $G$ is isomorphic to $mathbb{R}^2$ then you can precompose any isomorphism $mathbb{R}^2 mapsto G$ with any automorphism of $mathbb{R}^2$ to get a different isomorphism $mathbb{R}^2 mapsto G$.
$endgroup$
– Lee Mosher
Dec 23 '16 at 0:08
add a comment |
2 Answers
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active
oldest
votes
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I think you've got the intuition right.
When we have two isomorphic groups, we know that they're different as SETS, typically, but that doesn't really matter -- for all the algebraic things we'll look at (how many subgroups? abelian or not?) they're effectively "the same".
The same holds with X-morphic objects in most parts of mathematics: homeomorphic topological spaces, isomorphic vector spaces, etc.
For smooth manifolds, there are a bunch of constructions/computations that work purely on the "manifoldness" of the objects, and these are the ones that are invariant under diffeomorphism.
My one suggestion is that this is the right notion of "sameness" in differential topology rather than in differential geometry, which was the tag you included; for differential geometry, the right notion is "diffeomorphic in a way what the diffeormorphism makes the metrics on the domain and codomain correspond", i.e.
$$
langle u, v rangle_{M_p} =
langle dphi(p)(u), dphi(p)(v)rangle_{N_{phi(p)}}.
$$
where $phi:M to N$ is the diffeomorphism, and $langle,rangle_{M_p}$ means the inner product on the tangent space to $M$ at the point $p in M$.
By the way, the other answer says "not all of these things are true," but the claim about "same curvature" is true if taken in the right context: if you have a diffeomorphism from $M$ to $N$, and put a metric on $M$, then there's a metric on $N$ implicitly defined by the correspondence in the equation above. And the curvature of $N$ (with this particular metric!) at some point $phi(p)$ will be the same as the curvature of $M$ at $p$.
On the other hand, if $M$ and $N$ already have metrics and you build a diffeomorphism, there's no guarantee that the diffeomorphism will carry the metric on $M$ to the one on $N$, so any statements about curvature in this case are likely to be wrong. Example: a unit 2-sphere in 3-space has a metric inherited from that of 3-space; so does the radius-5 2-sphere in 3-space (again with the inherited metric), and there's an obvious diffeomorphism between them (scale up by 5)...but it doesn't carry one metric to the other, so the curvatures at corresponding points are rather different.
answered Dec 22 '16 at 21:55
John HughesJohn Hughes
64.4k24191
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add a comment |
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Not all of these are true. A diffeomorphism is essentially a bijective map that identifies one smooth structure with another, and therefore everything that can be defined purely in terms of these smooth structures should also be carried over. So, for instance, a globally defined vector field $X$ on $M$ can be pushed forward by a diffeomorphism $f:Mto N$ to a globally defined vector field $f_*X$ on $N$. (Note that pushforwards of vector fields are always defined locally, but not globally unless the map is a diffeomorphism.)
However, other than respecting smooth structures, a diffeomorphism can still be geometrically quite strange. In particular, a diffeomorphism doesn't need to respect anything regarding lengths or angles - the concepts that govern what we think of as how the manifold is actually shaped. The transformations that do respect these have to respect some additional structure, like a Riemannian metric, and give rise to more specialized diffeomorphisms like Riemannian isometries and conformal equivalences. In particular your example 4 is not correct: a diffeomorphism need not preserve curvature.
For an example, think of $M = S^2$, the standard unit sphere, and $N$ as an ellipsoid with axis lengths $a,b,c$, with at least two of the three axes being different. Then $M$ has constant sectional curvature and $N$ does not. In the category of smooth manifolds, we cannot tell $M$ and $N$ apart, since they are isomorphic in this category. But they are not isomorphic in the sense of Riemmanian isometry, so in the category of Riemannian manifolds they are distinct. That's the distinction between diffeomorphism and isometry: do you care about being able to distinguish between $M$ and $N$?
Intuitively, if I ask you to draw a diffeomorphism from a sphere to another $2$-manifold, you have a lot of freedom: any smooth genus $0$ oriented surface you draw will do the trick, and there's a lot of ways to draw those. But if I ask you to draw a Riemannian isometry, or a conformal equivalence, you have much less freedom: in the first case, you basically have to reproduce the original sphsere exactly, and in the second you get to scale the sphere but not much else.
answered Dec 22 '16 at 21:58
Gyu Eun LeeGyu Eun Lee
13.2k2353
$endgroup$
$begingroup$
Actually, $N$ has zero (intrinsic) curvature and they are isometric as Riemannian manifolds. It is true that $N$ has non-zero extrinsic curvature (as a curve in $mathbb{R}^2$) but the Riemannian structure doesn't see it.
$endgroup$
– levap
Dec 22 '16 at 22:07
$begingroup$
@levap Fixed, with new examples.
$endgroup$
– Gyu Eun Lee
Dec 22 '16 at 22:21
$begingroup$
Cool. The sphere example is good but regarding the last paragraph, again, the standard circle (with the standard Riemannian metric) will be isometric to any smooth simple closed loop of the same length so a circle can be isometric to an ellipse (as Riemannian manifolds).
$endgroup$
– levap
Dec 22 '16 at 22:30
$begingroup$
Aak, these darned $1$-manifolds! Thanks, fixed.
$endgroup$
– Gyu Eun Lee
Dec 22 '16 at 22:33
add a comment |
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2 Answers
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2 Answers
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$begingroup$
I think you've got the intuition right.
When we have two isomorphic groups, we know that they're different as SETS, typically, but that doesn't really matter -- for all the algebraic things we'll look at (how many subgroups? abelian or not?) they're effectively "the same".
The same holds with X-morphic objects in most parts of mathematics: homeomorphic topological spaces, isomorphic vector spaces, etc.
For smooth manifolds, there are a bunch of constructions/computations that work purely on the "manifoldness" of the objects, and these are the ones that are invariant under diffeomorphism.
My one suggestion is that this is the right notion of "sameness" in differential topology rather than in differential geometry, which was the tag you included; for differential geometry, the right notion is "diffeomorphic in a way what the diffeormorphism makes the metrics on the domain and codomain correspond", i.e.
$$
langle u, v rangle_{M_p} =
langle dphi(p)(u), dphi(p)(v)rangle_{N_{phi(p)}}.
$$
where $phi:M to N$ is the diffeomorphism, and $langle,rangle_{M_p}$ means the inner product on the tangent space to $M$ at the point $p in M$.
By the way, the other answer says "not all of these things are true," but the claim about "same curvature" is true if taken in the right context: if you have a diffeomorphism from $M$ to $N$, and put a metric on $M$, then there's a metric on $N$ implicitly defined by the correspondence in the equation above. And the curvature of $N$ (with this particular metric!) at some point $phi(p)$ will be the same as the curvature of $M$ at $p$.
On the other hand, if $M$ and $N$ already have metrics and you build a diffeomorphism, there's no guarantee that the diffeomorphism will carry the metric on $M$ to the one on $N$, so any statements about curvature in this case are likely to be wrong. Example: a unit 2-sphere in 3-space has a metric inherited from that of 3-space; so does the radius-5 2-sphere in 3-space (again with the inherited metric), and there's an obvious diffeomorphism between them (scale up by 5)...but it doesn't carry one metric to the other, so the curvatures at corresponding points are rather different.
answered Dec 22 '16 at 21:55
John HughesJohn Hughes
64.4k24191
$endgroup$
add a comment |
$begingroup$
I think you've got the intuition right.
When we have two isomorphic groups, we know that they're different as SETS, typically, but that doesn't really matter -- for all the algebraic things we'll look at (how many subgroups? abelian or not?) they're effectively "the same".
The same holds with X-morphic objects in most parts of mathematics: homeomorphic topological spaces, isomorphic vector spaces, etc.
For smooth manifolds, there are a bunch of constructions/computations that work purely on the "manifoldness" of the objects, and these are the ones that are invariant under diffeomorphism.
My one suggestion is that this is the right notion of "sameness" in differential topology rather than in differential geometry, which was the tag you included; for differential geometry, the right notion is "diffeomorphic in a way what the diffeormorphism makes the metrics on the domain and codomain correspond", i.e.
$$
langle u, v rangle_{M_p} =
langle dphi(p)(u), dphi(p)(v)rangle_{N_{phi(p)}}.
$$
where $phi:M to N$ is the diffeomorphism, and $langle,rangle_{M_p}$ means the inner product on the tangent space to $M$ at the point $p in M$.
By the way, the other answer says "not all of these things are true," but the claim about "same curvature" is true if taken in the right context: if you have a diffeomorphism from $M$ to $N$, and put a metric on $M$, then there's a metric on $N$ implicitly defined by the correspondence in the equation above. And the curvature of $N$ (with this particular metric!) at some point $phi(p)$ will be the same as the curvature of $M$ at $p$.
On the other hand, if $M$ and $N$ already have metrics and you build a diffeomorphism, there's no guarantee that the diffeomorphism will carry the metric on $M$ to the one on $N$, so any statements about curvature in this case are likely to be wrong. Example: a unit 2-sphere in 3-space has a metric inherited from that of 3-space; so does the radius-5 2-sphere in 3-space (again with the inherited metric), and there's an obvious diffeomorphism between them (scale up by 5)...but it doesn't carry one metric to the other, so the curvatures at corresponding points are rather different.
answered Dec 22 '16 at 21:55
John HughesJohn Hughes
64.4k24191
$endgroup$
add a comment |
$begingroup$
I think you've got the intuition right.
When we have two isomorphic groups, we know that they're different as SETS, typically, but that doesn't really matter -- for all the algebraic things we'll look at (how many subgroups? abelian or not?) they're effectively "the same".
The same holds with X-morphic objects in most parts of mathematics: homeomorphic topological spaces, isomorphic vector spaces, etc.
For smooth manifolds, there are a bunch of constructions/computations that work purely on the "manifoldness" of the objects, and these are the ones that are invariant under diffeomorphism.
My one suggestion is that this is the right notion of "sameness" in differential topology rather than in differential geometry, which was the tag you included; for differential geometry, the right notion is "diffeomorphic in a way what the diffeormorphism makes the metrics on the domain and codomain correspond", i.e.
$$
langle u, v rangle_{M_p} =
langle dphi(p)(u), dphi(p)(v)rangle_{N_{phi(p)}}.
$$
where $phi:M to N$ is the diffeomorphism, and $langle,rangle_{M_p}$ means the inner product on the tangent space to $M$ at the point $p in M$.
By the way, the other answer says "not all of these things are true," but the claim about "same curvature" is true if taken in the right context: if you have a diffeomorphism from $M$ to $N$, and put a metric on $M$, then there's a metric on $N$ implicitly defined by the correspondence in the equation above. And the curvature of $N$ (with this particular metric!) at some point $phi(p)$ will be the same as the curvature of $M$ at $p$.
On the other hand, if $M$ and $N$ already have metrics and you build a diffeomorphism, there's no guarantee that the diffeomorphism will carry the metric on $M$ to the one on $N$, so any statements about curvature in this case are likely to be wrong. Example: a unit 2-sphere in 3-space has a metric inherited from that of 3-space; so does the radius-5 2-sphere in 3-space (again with the inherited metric), and there's an obvious diffeomorphism between them (scale up by 5)...but it doesn't carry one metric to the other, so the curvatures at corresponding points are rather different.
answered Dec 22 '16 at 21:55
John HughesJohn Hughes
64.4k24191
$endgroup$
I think you've got the intuition right.
When we have two isomorphic groups, we know that they're different as SETS, typically, but that doesn't really matter -- for all the algebraic things we'll look at (how many subgroups? abelian or not?) they're effectively "the same".
The same holds with X-morphic objects in most parts of mathematics: homeomorphic topological spaces, isomorphic vector spaces, etc.
For smooth manifolds, there are a bunch of constructions/computations that work purely on the "manifoldness" of the objects, and these are the ones that are invariant under diffeomorphism.
My one suggestion is that this is the right notion of "sameness" in differential topology rather than in differential geometry, which was the tag you included; for differential geometry, the right notion is "diffeomorphic in a way what the diffeormorphism makes the metrics on the domain and codomain correspond", i.e.
$$
langle u, v rangle_{M_p} =
langle dphi(p)(u), dphi(p)(v)rangle_{N_{phi(p)}}.
$$
where $phi:M to N$ is the diffeomorphism, and $langle,rangle_{M_p}$ means the inner product on the tangent space to $M$ at the point $p in M$.
By the way, the other answer says "not all of these things are true," but the claim about "same curvature" is true if taken in the right context: if you have a diffeomorphism from $M$ to $N$, and put a metric on $M$, then there's a metric on $N$ implicitly defined by the correspondence in the equation above. And the curvature of $N$ (with this particular metric!) at some point $phi(p)$ will be the same as the curvature of $M$ at $p$.
On the other hand, if $M$ and $N$ already have metrics and you build a diffeomorphism, there's no guarantee that the diffeomorphism will carry the metric on $M$ to the one on $N$, so any statements about curvature in this case are likely to be wrong. Example: a unit 2-sphere in 3-space has a metric inherited from that of 3-space; so does the radius-5 2-sphere in 3-space (again with the inherited metric), and there's an obvious diffeomorphism between them (scale up by 5)...but it doesn't carry one metric to the other, so the curvatures at corresponding points are rather different.
answered Dec 22 '16 at 21:55
John HughesJohn Hughes
64.4k24191
edited Jan 20 at 16:15
answered Dec 22 '16 at 21:55
John HughesJohn Hughes
64.4k24191
answered Dec 22 '16 at 21:55
John HughesJohn Hughes
64.4k24191
answered Dec 22 '16 at 21:55
John HughesJohn Hughes
64.4k24191
64.4k24191
add a comment |
add a comment |
$begingroup$
Not all of these are true. A diffeomorphism is essentially a bijective map that identifies one smooth structure with another, and therefore everything that can be defined purely in terms of these smooth structures should also be carried over. So, for instance, a globally defined vector field $X$ on $M$ can be pushed forward by a diffeomorphism $f:Mto N$ to a globally defined vector field $f_*X$ on $N$. (Note that pushforwards of vector fields are always defined locally, but not globally unless the map is a diffeomorphism.)
However, other than respecting smooth structures, a diffeomorphism can still be geometrically quite strange. In particular, a diffeomorphism doesn't need to respect anything regarding lengths or angles - the concepts that govern what we think of as how the manifold is actually shaped. The transformations that do respect these have to respect some additional structure, like a Riemannian metric, and give rise to more specialized diffeomorphisms like Riemannian isometries and conformal equivalences. In particular your example 4 is not correct: a diffeomorphism need not preserve curvature.
For an example, think of $M = S^2$, the standard unit sphere, and $N$ as an ellipsoid with axis lengths $a,b,c$, with at least two of the three axes being different. Then $M$ has constant sectional curvature and $N$ does not. In the category of smooth manifolds, we cannot tell $M$ and $N$ apart, since they are isomorphic in this category. But they are not isomorphic in the sense of Riemmanian isometry, so in the category of Riemannian manifolds they are distinct. That's the distinction between diffeomorphism and isometry: do you care about being able to distinguish between $M$ and $N$?
Intuitively, if I ask you to draw a diffeomorphism from a sphere to another $2$-manifold, you have a lot of freedom: any smooth genus $0$ oriented surface you draw will do the trick, and there's a lot of ways to draw those. But if I ask you to draw a Riemannian isometry, or a conformal equivalence, you have much less freedom: in the first case, you basically have to reproduce the original sphsere exactly, and in the second you get to scale the sphere but not much else.
answered Dec 22 '16 at 21:58
Gyu Eun LeeGyu Eun Lee
13.2k2353
$endgroup$
$begingroup$
Actually, $N$ has zero (intrinsic) curvature and they are isometric as Riemannian manifolds. It is true that $N$ has non-zero extrinsic curvature (as a curve in $mathbb{R}^2$) but the Riemannian structure doesn't see it.
$endgroup$
– levap
Dec 22 '16 at 22:07
$begingroup$
@levap Fixed, with new examples.
$endgroup$
– Gyu Eun Lee
Dec 22 '16 at 22:21
$begingroup$
Cool. The sphere example is good but regarding the last paragraph, again, the standard circle (with the standard Riemannian metric) will be isometric to any smooth simple closed loop of the same length so a circle can be isometric to an ellipse (as Riemannian manifolds).
$endgroup$
– levap
Dec 22 '16 at 22:30
$begingroup$
Aak, these darned $1$-manifolds! Thanks, fixed.
$endgroup$
– Gyu Eun Lee
Dec 22 '16 at 22:33
add a comment |
$begingroup$
Not all of these are true. A diffeomorphism is essentially a bijective map that identifies one smooth structure with another, and therefore everything that can be defined purely in terms of these smooth structures should also be carried over. So, for instance, a globally defined vector field $X$ on $M$ can be pushed forward by a diffeomorphism $f:Mto N$ to a globally defined vector field $f_*X$ on $N$. (Note that pushforwards of vector fields are always defined locally, but not globally unless the map is a diffeomorphism.)
However, other than respecting smooth structures, a diffeomorphism can still be geometrically quite strange. In particular, a diffeomorphism doesn't need to respect anything regarding lengths or angles - the concepts that govern what we think of as how the manifold is actually shaped. The transformations that do respect these have to respect some additional structure, like a Riemannian metric, and give rise to more specialized diffeomorphisms like Riemannian isometries and conformal equivalences. In particular your example 4 is not correct: a diffeomorphism need not preserve curvature.
For an example, think of $M = S^2$, the standard unit sphere, and $N$ as an ellipsoid with axis lengths $a,b,c$, with at least two of the three axes being different. Then $M$ has constant sectional curvature and $N$ does not. In the category of smooth manifolds, we cannot tell $M$ and $N$ apart, since they are isomorphic in this category. But they are not isomorphic in the sense of Riemmanian isometry, so in the category of Riemannian manifolds they are distinct. That's the distinction between diffeomorphism and isometry: do you care about being able to distinguish between $M$ and $N$?
Intuitively, if I ask you to draw a diffeomorphism from a sphere to another $2$-manifold, you have a lot of freedom: any smooth genus $0$ oriented surface you draw will do the trick, and there's a lot of ways to draw those. But if I ask you to draw a Riemannian isometry, or a conformal equivalence, you have much less freedom: in the first case, you basically have to reproduce the original sphsere exactly, and in the second you get to scale the sphere but not much else.
answered Dec 22 '16 at 21:58
Gyu Eun LeeGyu Eun Lee
13.2k2353
$endgroup$
$begingroup$
Actually, $N$ has zero (intrinsic) curvature and they are isometric as Riemannian manifolds. It is true that $N$ has non-zero extrinsic curvature (as a curve in $mathbb{R}^2$) but the Riemannian structure doesn't see it.
$endgroup$
– levap
Dec 22 '16 at 22:07
$begingroup$
@levap Fixed, with new examples.
$endgroup$
– Gyu Eun Lee
Dec 22 '16 at 22:21
$begingroup$
Cool. The sphere example is good but regarding the last paragraph, again, the standard circle (with the standard Riemannian metric) will be isometric to any smooth simple closed loop of the same length so a circle can be isometric to an ellipse (as Riemannian manifolds).
$endgroup$
– levap
Dec 22 '16 at 22:30
$begingroup$
Aak, these darned $1$-manifolds! Thanks, fixed.
$endgroup$
– Gyu Eun Lee
Dec 22 '16 at 22:33
add a comment |
$begingroup$
Not all of these are true. A diffeomorphism is essentially a bijective map that identifies one smooth structure with another, and therefore everything that can be defined purely in terms of these smooth structures should also be carried over. So, for instance, a globally defined vector field $X$ on $M$ can be pushed forward by a diffeomorphism $f:Mto N$ to a globally defined vector field $f_*X$ on $N$. (Note that pushforwards of vector fields are always defined locally, but not globally unless the map is a diffeomorphism.)
However, other than respecting smooth structures, a diffeomorphism can still be geometrically quite strange. In particular, a diffeomorphism doesn't need to respect anything regarding lengths or angles - the concepts that govern what we think of as how the manifold is actually shaped. The transformations that do respect these have to respect some additional structure, like a Riemannian metric, and give rise to more specialized diffeomorphisms like Riemannian isometries and conformal equivalences. In particular your example 4 is not correct: a diffeomorphism need not preserve curvature.
For an example, think of $M = S^2$, the standard unit sphere, and $N$ as an ellipsoid with axis lengths $a,b,c$, with at least two of the three axes being different. Then $M$ has constant sectional curvature and $N$ does not. In the category of smooth manifolds, we cannot tell $M$ and $N$ apart, since they are isomorphic in this category. But they are not isomorphic in the sense of Riemmanian isometry, so in the category of Riemannian manifolds they are distinct. That's the distinction between diffeomorphism and isometry: do you care about being able to distinguish between $M$ and $N$?
Intuitively, if I ask you to draw a diffeomorphism from a sphere to another $2$-manifold, you have a lot of freedom: any smooth genus $0$ oriented surface you draw will do the trick, and there's a lot of ways to draw those. But if I ask you to draw a Riemannian isometry, or a conformal equivalence, you have much less freedom: in the first case, you basically have to reproduce the original sphsere exactly, and in the second you get to scale the sphere but not much else.
answered Dec 22 '16 at 21:58
Gyu Eun LeeGyu Eun Lee
13.2k2353
$endgroup$
Not all of these are true. A diffeomorphism is essentially a bijective map that identifies one smooth structure with another, and therefore everything that can be defined purely in terms of these smooth structures should also be carried over. So, for instance, a globally defined vector field $X$ on $M$ can be pushed forward by a diffeomorphism $f:Mto N$ to a globally defined vector field $f_*X$ on $N$. (Note that pushforwards of vector fields are always defined locally, but not globally unless the map is a diffeomorphism.)
However, other than respecting smooth structures, a diffeomorphism can still be geometrically quite strange. In particular, a diffeomorphism doesn't need to respect anything regarding lengths or angles - the concepts that govern what we think of as how the manifold is actually shaped. The transformations that do respect these have to respect some additional structure, like a Riemannian metric, and give rise to more specialized diffeomorphisms like Riemannian isometries and conformal equivalences. In particular your example 4 is not correct: a diffeomorphism need not preserve curvature.
For an example, think of $M = S^2$, the standard unit sphere, and $N$ as an ellipsoid with axis lengths $a,b,c$, with at least two of the three axes being different. Then $M$ has constant sectional curvature and $N$ does not. In the category of smooth manifolds, we cannot tell $M$ and $N$ apart, since they are isomorphic in this category. But they are not isomorphic in the sense of Riemmanian isometry, so in the category of Riemannian manifolds they are distinct. That's the distinction between diffeomorphism and isometry: do you care about being able to distinguish between $M$ and $N$?
Intuitively, if I ask you to draw a diffeomorphism from a sphere to another $2$-manifold, you have a lot of freedom: any smooth genus $0$ oriented surface you draw will do the trick, and there's a lot of ways to draw those. But if I ask you to draw a Riemannian isometry, or a conformal equivalence, you have much less freedom: in the first case, you basically have to reproduce the original sphsere exactly, and in the second you get to scale the sphere but not much else.
answered Dec 22 '16 at 21:58
Gyu Eun LeeGyu Eun Lee
13.2k2353
edited Dec 22 '16 at 22:32
answered Dec 22 '16 at 21:58
Gyu Eun LeeGyu Eun Lee
13.2k2353
answered Dec 22 '16 at 21:58
Gyu Eun LeeGyu Eun Lee
13.2k2353
answered Dec 22 '16 at 21:58
Gyu Eun LeeGyu Eun Lee
13.2k2353
13.2k2353
$begingroup$
Actually, $N$ has zero (intrinsic) curvature and they are isometric as Riemannian manifolds. It is true that $N$ has non-zero extrinsic curvature (as a curve in $mathbb{R}^2$) but the Riemannian structure doesn't see it.
$endgroup$
– levap
Dec 22 '16 at 22:07
$begingroup$
@levap Fixed, with new examples.
$endgroup$
– Gyu Eun Lee
Dec 22 '16 at 22:21
$begingroup$
Cool. The sphere example is good but regarding the last paragraph, again, the standard circle (with the standard Riemannian metric) will be isometric to any smooth simple closed loop of the same length so a circle can be isometric to an ellipse (as Riemannian manifolds).
$endgroup$
– levap
Dec 22 '16 at 22:30
$begingroup$
Aak, these darned $1$-manifolds! Thanks, fixed.
$endgroup$
– Gyu Eun Lee
Dec 22 '16 at 22:33
add a comment |
$begingroup$
Actually, $N$ has zero (intrinsic) curvature and they are isometric as Riemannian manifolds. It is true that $N$ has non-zero extrinsic curvature (as a curve in $mathbb{R}^2$) but the Riemannian structure doesn't see it.
$endgroup$
– levap
Dec 22 '16 at 22:07
$begingroup$
@levap Fixed, with new examples.
$endgroup$
– Gyu Eun Lee
Dec 22 '16 at 22:21
$begingroup$
Cool. The sphere example is good but regarding the last paragraph, again, the standard circle (with the standard Riemannian metric) will be isometric to any smooth simple closed loop of the same length so a circle can be isometric to an ellipse (as Riemannian manifolds).
$endgroup$
– levap
Dec 22 '16 at 22:30
$begingroup$
Aak, these darned $1$-manifolds! Thanks, fixed.
$endgroup$
– Gyu Eun Lee
Dec 22 '16 at 22:33
$begingroup$
Actually, $N$ has zero (intrinsic) curvature and they are isometric as Riemannian manifolds. It is true that $N$ has non-zero extrinsic curvature (as a curve in $mathbb{R}^2$) but the Riemannian structure doesn't see it.
$endgroup$
– levap
Dec 22 '16 at 22:07
$begingroup$
Actually, $N$ has zero (intrinsic) curvature and they are isometric as Riemannian manifolds. It is true that $N$ has non-zero extrinsic curvature (as a curve in $mathbb{R}^2$) but the Riemannian structure doesn't see it.
$endgroup$
– levap
Dec 22 '16 at 22:07
$begingroup$
@levap Fixed, with new examples.
$endgroup$
– Gyu Eun Lee
Dec 22 '16 at 22:21
$begingroup$
@levap Fixed, with new examples.
$endgroup$
– Gyu Eun Lee
Dec 22 '16 at 22:21
$begingroup$
Cool. The sphere example is good but regarding the last paragraph, again, the standard circle (with the standard Riemannian metric) will be isometric to any smooth simple closed loop of the same length so a circle can be isometric to an ellipse (as Riemannian manifolds).
$endgroup$
– levap
Dec 22 '16 at 22:30
$begingroup$
Cool. The sphere example is good but regarding the last paragraph, again, the standard circle (with the standard Riemannian metric) will be isometric to any smooth simple closed loop of the same length so a circle can be isometric to an ellipse (as Riemannian manifolds).
$endgroup$
– levap
Dec 22 '16 at 22:30
$begingroup$
Aak, these darned $1$-manifolds! Thanks, fixed.
$endgroup$
– Gyu Eun Lee
Dec 22 '16 at 22:33
$begingroup$
Aak, these darned $1$-manifolds! Thanks, fixed.
$endgroup$
– Gyu Eun Lee
Dec 22 '16 at 22:33
add a comment |
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$begingroup$
To see you write that you "have no trouble thinking" of two isomorphic groups as "exactly the same thing up to a (superficial) renaming of the elements" tells me that you misunderstand the nature of isomorphism. The renaming is far from superficial, given that it is not unique. For example, the automorphism group of $mathbb{R}^2$ under addition is uncountably infinite, and if $G$ is isomorphic to $mathbb{R}^2$ then you can precompose any isomorphism $mathbb{R}^2 mapsto G$ with any automorphism of $mathbb{R}^2$ to get a different isomorphism $mathbb{R}^2 mapsto G$.
$endgroup$
– Lee Mosher
Dec 23 '16 at 0:08